CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

An object of mass 0.2 kg executes simple harmonic motion along X-axis with frequency of 25/ π Hz. At the position x = 0.04 m, the object has kinetic energy of 0.5 J and potential energy of 0.4 J. The amplitude of oscillation in meter is equal to


A

0.05 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

0.06 m

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C

0.01 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

0.002 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B

0.06 m


E=12mω2A2

E=12m(2πf)2A2

A=12πf2Em

Putting E = K + U we obtain, A=12π(25π)2×(0.5+0.4)0.2

A=0.06m.


flag
Suggest Corrections
thumbs-up
7
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Energy in SHM
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon